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Assume we have $X, Y$ constant unit vectors of $\mathbb{R}^3$

I postulate that the maximum of the function:

$(V \cdot X) (V \cdot Y)$

I reached by the halfway vector between $X,Y$ i.e at the vector $V_0 = slerp(X,Y, 0.5)$

To try to prove it I tried finding the critical point of the derivative, i.e:

$(V'\cdot X)(V\cdot Y) + (V\cdot X)(V'\cdot Y)$

But that is leading me down a rabbit hole I don't seem to be able to get out of.

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  • $\begingroup$ Derivative with respect to $V$? But you'd better constrain $V$ to be a unit vector as well. $\endgroup$ – Ted Shifrin Jun 5 '20 at 3:50
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Without loss of generality, choose $X=(1,0,0)$ and $Y=(\cos\phi_0,\sin\phi_0,0)$. Then (assuming $V$ is also unit vector), you can write $V$ in polar coordinates as $V=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$. Then your expression becomes $$\sin\theta\cos\phi(\sin\theta\cos\phi\cos\phi_0+\sin\theta\sin\phi\sin\phi_0)=\sin^2\theta\cos\phi\cos(\phi-\phi_0)$$ If you want the maximum, you get $\theta=\pi/2$, so it's in the same plane. Also $$\frac d{d\phi}\cos\phi\cos(\phi-\phi_0)=-\sin(2\phi-\phi_0)=0$$ so $$\phi=\frac{\phi_0}2$$ You will need to consider separately the case where $\phi_0=\pi$

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